Download Chapter 10, Part 2

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
December 9, 2009
•F i n a l E x a m R e m i n d e r s :
•Date:
Monday, December 14, 2009
•Time: 8:00-10:30am
•Place: IRC 1 (Here)
•What to Bring:
Calculator
#2 Pencil
2 Pages (double-sided) of notes
•M a k e u p L a b - O n O W L , d u e a t f i n a l e x a m
•T o d a y
•Chapter
10: Gases
•F r i d a y
•Chapter
11: Intermolecular Forces and the Liquid State (Fri.)
Kinetic Molecular Theory & Gases
 Gas molecules far apart, always in motion, collide
with walls of container (pressure)
 Temperature  Average Kinetic Energy

Higher temperature = higher average kinetic energy
 Kinetic energy and velocity are not the same

Higher molar mass will move slower at the same temperature
 Boltzmann distribution
Plot of molecular speed (x) vs. number of molecules (y)
 Shows range of speeds in a collection of molecules

Gases- Equations from last time…
 Kinetic energy of one molecule
1
2
KE

mv
where
m

mass
,
v

veloc
2
 Average Kinetic energy of many molecules
12 3
KE

m
v

RT
2 2
T

temperatur
e
in
Kelvin
,
8.3145
J
0.082057
L

atm
R

gas
constant


mol

K
K

mol
 Average speed of molecules
3
RT
2
v

v


8
.
3415
J
/
K

mol
,
M

mol
ma
in
kg
/
m
rms R
M
Gas Diffusion
Gas Effusion
 Graham’s Law of Effusion
Graham’s Law Example
A sample of ethane, C2H6, effuses through a small hole
at a rate of 3.6 x 10-6 mol/hr. An unknown gas, under
the same conditions, effuses at a rate of 1.3 x 10-6
mol/hr. Calculate the molar mass of the unknown gas.
Back to Pressure…
 Collisions between gas molecules and container
exerts a force on container wall

More collisions and more energetic collisions = greater force =
higher pressure
 Kinetic molecular theory gives conceptual framework
for understanding the behavior of gases


P and n
P and T
(P  n)
(P  T)
Today’s Temp: 35°F
Pressure
Gauge
Today’s Temp: 85°F
Pressure
Gauge
Back to Pressure…
 Collisions between gas molecules and container
exerts a force on container wall

More collisions and more energetic collisions = greater force =
higher pressure
 Kinetic molecular theory gives conceptual framework
for understanding the behavior of gases




P and n
P and T
V and T
P and V
(P  n)
(P  T)
(V  1/T)
(P  1/V)
Volume of balloon at
room temperature
Volume of balloon at
5°C
The Gas Laws
 Boyle’s
P1V1  P2V2
Ideal Gas Law
PV  nRT
 Charles’
V1 V2

T1 T2
 Avogadro’s
V1 V2

n1 n2
R  0.082057 L  atm K  mol

Using the Gas Laws
 If you know 3 variables, solve for the fourth
 If some properties are constant, trends in one
property can predict another
 STP (“Standard Temperature and Pressure”)
T= 273 K
 P= 1 atm

 Strategy: Take note of the variables you know, find
the appropriate equation, then “plug and chug”
(or just remember PV=nRT and derive the appropriate
equation every time)
Examples
1.
A gas has a volume of 3L at 2 atm. What is its
volume at 4 atm?
2. A gas has a volume of 4.1 L at 127 C. What is its
volume at 227?
Examples
3. What volume does 7.4 grams of ethane (C2H2)
occupy at standard temperature and pressure?
4. The propane tank of a camping stove contains
3,000g of liquid C3H8. How large a container would
be needed to hold the same amount of propane as a
gas at 25C and 2250mmHg?
Related documents